l12 optim constrain

7/30/2019 L12 Optim Constrain

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Constrained Optimization

- Linear Programming

- Nonlinear Constrained Optimization

- MATLABs fmincon Function

Numerical Methods forNumerical Methods forCivilCivilEngineersEngineers

Lecture 12

Mongkol JIRAVACHARADETSchool ofCivil Engineering

S u r a n a r e eUNIVERSITY OF TECHNOLOGY

NAKORNRATCHSIMA


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LINEAR PROGRAMMINGStandard Form

Objective Function: MaximizeZ= c1x1+ c2x2+ . . . + cnxn

where cj = payoff ofjth activity

xj = magnitude ofjth activity

Z= total payoff due to all activities

Constraints: ai1x1 + ai2x2 + . . . + ainxn bi

where aij = amount ofith resource used byjth activity

bi = amount ofith resource available


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Example: Setting Up the Linear Programming Problem

Producing product with limited resources: Gas processing plant

Resource

Product

Regular Premium

Resource

Availability

Raw gas

Production time

Storage

Profit

7 m3/ton 11 m3/ton

10 hr/ton 8 hr/ton

9 ton 6 ton

150 / ton 175 / ton

77 m3/week

80 hr/week

Set x1 = amount of regular gas produced

x2 = amount of premium gas produced


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Total profit = 150x1

+ 175x2

MaximizeZ = 150x1 + 175x2 Objective function

Total gas used = 7x1 + 11x2

cannot exceed 77 m3/week

7x1 + 11x2 77 Constraints

10x1 + 8x2 80

x1 9 x2 6

x1

, x2

0


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10x1+8x2 = 80

7x1+11x2 = 77

11

7

7x1+11x2 = 77

11

7

9

6

10x1+8x2 80

7x1+11x2 77

Graphical Solution MaximizeZ = 150x1 + 175x2

1 2

1 2

1

2

7 11 77 (1)

10 8 80 (2)

0 9 (3)

0 6 (4)

x x

x x

x

x

+

+

x2

x10


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x2

x10

Increasing Objective Function From Z = 150x1

+ 175x2

ForZ= 0, 150x1 + 175x2 = 0

Z = 0 Z = 600

Z = 1400

Unique solution


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Other Possible Outcomes

x2

x10

Objective function parallel

to constraint.Alternate Solution


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x2

x10

No Feasible Solution


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x2

x10

Unbounded Problems

Z


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Simplex Method

Assumption: optimal solution will be an extreme point

Extreme points: corner pointswhere two constraints meet

x2

x10Z = 0 Z = 600

Z = 1400Objective function

isLINEAR


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Slack Variables

Reformulate constraint inequalities as equalities

1 2 1 2 1

7 11 77 7 11 77x x x x S+ + + =

slack variable

Resource constraint How muchslackof the resource

is available ?

IF . . .

S1

= + Surplus resource is not fully used.

S1 = - Exceeded the constraint.

S1 = 0 Exactly meet the constraint. All resource is used.


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Fully Augmented Version

MaximizeZ = 150x1 + 175x2

subject to

1 2

1 2

1

2

7 11 77

10 8 800 9

0 6

x x

x xx

x

+

+

1 2 1

1 2 2

1 3

2 4

1 2 1 2 3 4

7 11 77

10 8 809

6

, , , , , 0

x x S

x x Sx S

x S

x x S S S S

+ + =

+ + =

+ =

+ =

Linear algebraic equations

solving 4 equations with 6 unknowns


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Algebraic Solution

Underspecified: more unknowns (6) than equations (4)

Extreme Point Zero Variables

A x1, x2

B x2, S2

C S1, S2

D S1, S4

E x1, S4

x2

x1

x1 = 9, S3 = 0

9

6

0

10x1 + 8x2 = 80S2 = 0

7x1 + 11x2 = 77

S1 = 0

x2 = 6, S4 = 0

A B

C

DE

@ Extreme point :

4 equations with 4 unknowns

The problem can be solved.


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For pointE: Settingx1 = S4 = 0

The standard form reduced to

1 2 1

1 2 2

1 3

2 4

7 11 77

10 8 80

9

6

x x S

x x S

x S

x S

+ + =

+ + =

+ =

+ =

0

0

0

0

2 1

2 2

3

2

11 77

8 80

9

6

x S

x S

S

x

+ =

+ =

=

=

which can be solved for

x1 x2 S1 S2 S3 S4

0 6 11 32 9 0

nonbasic variables

basic variables


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Simplex Method Implementation

Start at pointA : Settingx1 = x2 = 0

1

2

3

4

77

80

9

6

S

S

S

S

=

=

=

=

Tableau: Z- 150x1 - 175x2 - 0S1 - 0S2 - 0S3 - 0S4 = 0

Basic Z x1 x2 S1 S2 S3 S4 Solution Intercept

Z

S1

S2

S3

S4

1

0

0

0

0

-150

7

10

1

0

-175

11

8

0

1

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0

77

80

9

6

11

8

9


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Graphical Depiction of Simplex Method

x2

x18

6

0A B

C

DE

S1

11F

S2

S4

S3

Changex1 to basic variable (entering variable)

Move fromA horizontally

Intercept(for each S)

=Solution

x1

coeff.

Check intercept:

S2 give min. intercept = 8

Therefore, change S2 to nonbasic(leaving) variable.


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Move to pointB : Settingx2 = S2 = 0 1 1

1

1 3

4

7 77

10 80

9

6

x S

x

x S

S

+ =

=

+ =

=

Basic Z x1 x2 S1 S2 S3 S4 Solution

Z

S1

x1

S3

S4

1

0

0

0

0

-150

7

1

1

0

-175

11

0.8

0

1

0

1

0

0

0

0

0

0.1

0

0

0

0

0

1

0

0

0

0

0

1

0

77

8

9

6

Pivot rowS2: Dividing row by 10 and replacing S2byx1

x -150

Z 1 0 -55 0 15 0 0 1200


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Z

S1x1

S3

S4

1

00

0

0

0

01

0

0

-55

5.40.8

-0.8

1

0

10

0

0

15

-0.70.1

-0.1

0

0

00

1

0

0

00

0

1

1200

218

1

6

Perform same operation on the remaining rows,

Intercept

3.88910

-1.25

6

- From A toB, objective function has increased to 1200

- S1 give min. positive intercept = 3.889 is entering variable

- Move fromB to C


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Z

S1x1

S3

S4

1

00

0

0

0

01

0

0

0

10

0

0

10.19

0.19-0.15

0.15

-0.19

7.87

-0.130.20

-0.20

0.13

0

00

1

0

0

00

0

1

1414

3.894.89

4.11

2.11

Final Tableau:

- No negative coeff. in objective func. row, No more increasing.

Final solution: x1 = 3.89, x2 = 4.89

Z= 1414


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MATLABs fmincon Function

( )1 2 21 2 1 2 2( ) 4 2 4 2 1xf x e x x x x x= + + + +

subject to the constraints:1 2 1 2

1 2

1.5

10

x x x x

x x

rewrite the constraints in the form c(x) 0 :

1 2 1 2

1 2

1.5 0

10 0

x x x x

x x

+


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Write M-file confun.m for the constraints

function [c, ceq] = confun(x)

% nonlinear inequality constraints

c = [1.5 + x(1)*x(2) - x(1) - x(2);

-x(1)*x(2) - 10];% nonlinear equality constraints

ceq = [];

> x0 = [-1,1];

> options = optimset(LargeScale,off);

> [x, fval] = . . .

fmincon(objfun,x0,[],[],[],[],[],[],confun,options)

x =

-9.5474 1.0474

fval =

0.0236

l12 optim constrain

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